waht - Thanks. And that part I understand just fine. There is only a z-component of the electric field, so when I do a cross product, I get both an r-component and a theta-component.
gabbagabbahey - So the way you're describing it, the magnetic field seems to come up at the center of the...
Homework Statement
The question actually asks for the equation for the magnetic field for the rotating disc; but all I'm after is the direction of the magnetic field.
Homework Equations
None were given; but I've been using:
\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}...
Ok, I have from my teacher that A = \frac{1}{3}, and that \left\langle S_{x}\right\rangle = \frac{2\hbar}{9}.
I can normalize it and get A = \frac{1}{3}; but I can't get the correct expectation value. Here is what I am doing:
\left\langle X | S_{x}X\right\rangle = A^{2}\frac{h}{2}[(1+2i)...
Any thoughts on normalizing this? I tried another way and got \frac{1}{3} and another way and I got \frac{16\hbar - 2\hbar i}{18}. And neither of those seem to work. I thought that 1/3 would work; but when I try to find the expectation value I get \frac{\hbar}{2} \frac{13}{3}.
I'm just not sure how to get from my relation: a = \mp ib to:
X^{y}_{+} = \frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ i \end{pmatrix}
X^{y}_{-} = \frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ -i \end{pmatrix}
Well, after a few days I'm still a bit confused with this problem. According to wikipedia the eigenspinors of S_{y} are:
X^{y}_{+} = \frac{1}{sqrt{2}}\left[ \begin{pmatrix}1 \\ i \end{pmatrix}\right]
X^{y}_{-} = \frac{1}{sqrt{2}}\left[ \begin{pmatrix}1 \\ -i \end{pmatrix}\right]...
[SOLVED] Electron Spin State and Values
Homework Statement
An electron is in the spin state:
X = A\begin{pmatrix} 1-2i \\ 2 \end{pmatrix}
(a) Determine the constant A by normalizing X
(b) If you measured S_{z} on this electron, what values could you get, and what is the probability...
[SOLVED] Expectation Values of Spin Operators
Homework Statement
b) Find the expectation values of S_{x}, S_{y}, and S_{z}
Homework Equations
From part a)
X = A \begin{pmatrix}3i \\ 4 \end{pmatrix}
Which was found to be: A = \frac{1}{5}
S_{x} = \begin{pmatrix}0 & 1 \\ 1 & 0...
Count Iblis - I understand what you mean. And that makes perfect sense too, that's a great way of getting the different spins. So wikipedia shows the S_{y} e-spinors as something different. They show:
S_{y}:
X^{y}_{+} = \frac{1}{\sqrt{2}} \left[ \begin{array}{cc}1 \\i...
nrqed - Right! Yeah, I didn't mean for my example of finding the eigenvalues as the problem that I was trying to solve. So...
\begin{pmatrix}-\lambda -i\frac{\hbar}{2} \\i\frac{\hbar}{2} -\lambda \end{pmatrix} = 0
\lambda^{2} =\frac{\hbar^{2}}{4} ---> \lambda = \pm \frac{\hbar}{2}...